I have the following system:



The problem asks for applying Euler's method and then finding an expression for the global error. Finally, supposing that

$$\lim_{h->0} \frac{1-(\frac{1+h}{e^h})^\frac{x}{h}}{\frac{x}{h}(e^h-(1+h))}=1$$

Check that the global error tends to $0$ as $h$ tends to $0$ as well.

Let's first apply Euler's method.




Since $x_i=ih$, we have $\eta_{i+1}=\eta_i+h(\eta_i + e^{ih}$). Factoring by $\eta_i$:


And now I don't how should I proceed. Should I solve the iteration? I've done problems where I can express $\eta_{i+1}$ in terms of $\eta_0$ by iterations, but here I can't do so because the $he^{ih}$ prevents that (or at least I haven't managed to do it).

I know that global error is defined as $e(x,h)=\eta(x,h)-y(x)$ but I don't know how exactly should I use that formula. My guess it that I have to solve the differential equation and use the solution $y(x)$, but I don't know how to solve that system or what exactly is $\eta(x,h)$.

Any help regarding to solving this problem will be highly appreciated.



The integrating factor for the exact solution is $e^{-x}$, so that $$ (e^{-x}y(x))'=1\implies y(x)=y(0)e^x+xe^x. $$ This resonance behavior will make the analysis of the numerical solution a little more complicated.

The Euler approximations have the general form $y_i=A(1+h)^i+Be^{ih}$. Inserting into the recursion one finds $$ Be^h=(1+h)B+h\implies B=\frac{h}{e^h-1-h} $$ and from the initial condition $$ y_0=A+B\implies y_i=(y_0-B)(1+h)^i+Be^{ih}=y_0(1+h)^i+\frac{h((1+h)^i-e^{ih})}{e^h-1-h} $$

Now you can insert the expansions for $$ (1+h)^i=\exp(i\ln(1+h))=\exp(ih-\tfrac12ih^2+\tfrac13ih^3+...) $$ to find the lower error order terms.of the difference to the exact solution.


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